Difference between revisions of "2012 AIME II Problems/Problem 7"
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Now we just proceed with the other 12 spaces with 7 1's, and we're looking for the <math>1000-495=505th</math> number. Well, <math>\binom{11}{7}=330</math>, so we know that the <math>2^{11}</math> digit also is 1, and we're left with finding the <math>505-330=175th</math> number with 11 spaces and 6 1's. Now <math>\binom{10}{6}=210,</math> which is too big, but <math>\binom{9}{6}=84.</math> Thus, the <math>2^9</math> digit is 1, and we're now looking for the <math>175-84=91st</math> number with 9 spaces and 5 1's. Continuing the same process, <math>\binom{8}{5}=56</math>, so the <math>2^8</math> digit is 1, and we're left to look for the <math>91-56=35th</math> number with 8 spaces and 4 1's. But here <math>\binom{7}{4}=35</math>, so N must be the last or largest 7-digit number with 4 1's. Thus the last 8 digits of <math>N</math> must be <math>01111000</math>, and to summarize, <math>N=1101101111000</math> in base <math>2</math>. Therefore, <math>N = 8+16+32+64+256+512+2048+4096 \equiv 32 \pmod{1000}</math>, and the answer is <math>\boxed{032}</math>. | Now we just proceed with the other 12 spaces with 7 1's, and we're looking for the <math>1000-495=505th</math> number. Well, <math>\binom{11}{7}=330</math>, so we know that the <math>2^{11}</math> digit also is 1, and we're left with finding the <math>505-330=175th</math> number with 11 spaces and 6 1's. Now <math>\binom{10}{6}=210,</math> which is too big, but <math>\binom{9}{6}=84.</math> Thus, the <math>2^9</math> digit is 1, and we're now looking for the <math>175-84=91st</math> number with 9 spaces and 5 1's. Continuing the same process, <math>\binom{8}{5}=56</math>, so the <math>2^8</math> digit is 1, and we're left to look for the <math>91-56=35th</math> number with 8 spaces and 4 1's. But here <math>\binom{7}{4}=35</math>, so N must be the last or largest 7-digit number with 4 1's. Thus the last 8 digits of <math>N</math> must be <math>01111000</math>, and to summarize, <math>N=1101101111000</math> in base <math>2</math>. Therefore, <math>N = 8+16+32+64+256+512+2048+4096 \equiv 32 \pmod{1000}</math>, and the answer is <math>\boxed{032}</math>. | ||
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+ | == Video Solution == | ||
+ | [https://youtu.be/d2UTg7OsYuE?si=-s4Ze-bsE2w-J5ij 2012 AIME II #12] | ||
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+ | [https://mathproblemsolvingskills.wordpress.com/ Math Problem Solving Skills] | ||
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== See Also == | == See Also == |
Revision as of 19:09, 19 July 2024
Contents
[hide]Problem 7
Let be the increasing sequence of positive integers whose binary representation has exactly ones. Let be the 1000th number in . Find the remainder when is divided by .
Solution
Okay, an exercise in counting (lots of binomials to calculate!). In base 2, the first number is , which is the only way to choose 8 1's out of 8 spaces, or . What about 9 spaces? Well, all told, there are , which includes the first 1. Similarly, for 10 spaces, there are which includes the first 9. For 11 spaces, there are , which includes the first 45. You're getting the handle. For 12 spaces, there are , which includes the first 165; for 13 spaces, there are , so we now know that has exactly 13 spaces, so the digit is 1.
Now we just proceed with the other 12 spaces with 7 1's, and we're looking for the number. Well, , so we know that the digit also is 1, and we're left with finding the number with 11 spaces and 6 1's. Now which is too big, but Thus, the digit is 1, and we're now looking for the number with 9 spaces and 5 1's. Continuing the same process, , so the digit is 1, and we're left to look for the number with 8 spaces and 4 1's. But here , so N must be the last or largest 7-digit number with 4 1's. Thus the last 8 digits of must be , and to summarize, in base . Therefore, , and the answer is .
Video Solution
See Also
Video Solution: https://www.youtube.com/watch?v=WRP--6Db1qw
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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